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A forest ranger has data on the heights of a large growth of young pine trees. The mean height is 3.2 feet and the standard deviation 0.6 feet. A histogram shows that the distribution of heights is approximately normal. Approxi mately what fraction of the trees should we expect to be between 4.0 and 4.4 feet tall? A) 2% B) 7 % C ) 9\% D 91% E) 98%

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johnlloydrone

Answer:

The mean height is 3.2 feet and thestandard deviation 0.6 feet. A… ... A forest ranger has data on the heights of a large growth ... mately what fraction of the trees should we expect to be.

raphealnwobi

The  fraction of the trees that is between 4.0 and 4.4 feet is 7%

The z score is used to determine by how many standard deviations the raw score is above or below the mean. The z score is given by:

[tex]z=\frac{x-\mu}{\sigma}[/tex]

x is the raw score, μ is mean and σ is standard deviation

Given that μ = 3.2, σ = 0.6, hence:

[tex]For\ x = 4:\\\\z=\frac{4-3.2}{0.6} =1.33\\\\\\For\ x = 4.4:\\\\z=\frac{4.4-3.2}{0.6} =2[/tex]

From the normal distribution table, P(4 < x < 4.4) = P(1.33 < z < 2) = P(z < 2) - P(z < 1.33) = 0.9772 - 0.9082 = 7%

The  fraction of the trees that is between 4.0 and 4.4 feet is 7%

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